Maths CE 1966
2010-11-25 6:02 am
A, B and C are three points on a circle. The perpendicular line from C to AB cuts AB at F and the circle at G. The perpendicular line from C to AB at F and the circle at G. The perpendicular line from A to BC cuts CG at O and BC at D. Prove that F is the mid-point of OG.
回答 (2)
2010-11-25 6:18 am
✔ 最佳答案
Join AG.Let angle AGC = a and angle AOG = b
angle BFO + angle BDO = 90 + 90 = 180
So B,D,O,F are concyclic. (opp. angles supp.)
Hence angle ABC = angle AOG = b (ext. angle, cyclic quad.)
But angle AGC = angle ABC = a (angles in the same segment)
So a = b
Hence AG = AO (sides opp. equal angles)
Since triangle AGO is isosceles and AF perpendicular OG,
OF = GF (property of isos. triangle)
Thus F is the mid-point of OG.
參考: me
2010-11-25 6:18 am
(1) Triangle AFO is similar to triangle ODC (AAA), so angle FAO = angle OCD.
(2) Angle GAF = angle GCB (angle in the same segment).
Therefore, angle GAF = angle FAO, so triangle AGF congruent triangle AFO (ASA), therefore, GF = FO, that means, F is the mid-point of OG.
(2) Angle GAF = angle GCB (angle in the same segment).
Therefore, angle GAF = angle FAO, so triangle AGF congruent triangle AFO (ASA), therefore, GF = FO, that means, F is the mid-point of OG.
收錄日期: 2021-04-25 22:38:56
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